# Step in Exact Differential Equations

I'm going through this "Elementary Differential Equations and Boundary Value Problems" book, by Boyce and DiPrima, and I have a little trouble understanding a passage in the Exact Equations explanation.

"Let $$\psi(x,y) = c$$ $$(1)$$ where c is a constant. Assuming that $$(1)$$ defines $$y$$ implicitly as a differentiable function of $$x$$, we can differentiate $$(1)$$ with respect to x and obtain $$\psi_{x}(x,y) + \psi_{y}(x,y)y' = 0$$." (ipsis litteris)

How differentiating with respect to $$x$$ leads up to the second equation exactly? Thanks for any input :)

Let $$f(x,z):= \psi(x,z)$$, $$g(x):=(x,y(x))$$ an $$h=f\circ g$$. Then, from the chain rule $$D h(x)= Df(g(x)) D g(x)=(\psi_x(x,y(x)) \, , \, \psi_y(x,(y(x))^\top (1 \, , \, y'(x)) = \psi_x(x,y(x)) + \psi_y(x,y(x))y'(x)$$ Since $$f$$ is constant, the derivative of $$h$$ is $$0$$, that is, we have $$\psi_x(x,y(x)) + \psi_y(x,y(x))y'(x) = 0.$$